Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1011
10
avatar+262 

=0

 Jul 13, 2015

Best Answer 

 #4
avatar+26396 
+10

  ax2+bx+c=0x1,2=b±b24ac2a  

 

a=m  b=1  c=m24x1,2=b±b24ac2ax1,2=1±124m(m24)2mx1,2=1±1+m(m2)2mx1,2=1±1+m22m2mx1,2=1±12m+m22mx1,2=1±(1m)22mx1,2=1±(1m)2mx1=1+(1m)2mx1=2m2mx1=22mm2mx1=1m12x2=1(1m)2mx2=11+m)2mx2=m2mx2=12

 

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 Jul 13, 2015
 #1
avatar+14538 
+5

mx2x(m2)4=0

 

integer solutions :   m = -2       x = 1

                               m = 2        x = 0

 Jul 13, 2015
 #2
avatar+33654 
+10

geno has already answered this here: http://web2.0calc.com/questions/help-me-please_38 

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 Jul 13, 2015
 #3
avatar+118696 
+5

Sabi92, if you do not understand Geno's answer, you need to post questions about it on the original thread. :)

 Jul 13, 2015
 #4
avatar+26396 
+10
Best Answer

  ax2+bx+c=0x1,2=b±b24ac2a  

 

a=m  b=1  c=m24x1,2=b±b24ac2ax1,2=1±124m(m24)2mx1,2=1±1+m(m2)2mx1,2=1±1+m22m2mx1,2=1±12m+m22mx1,2=1±(1m)22mx1,2=1±(1m)2mx1=1+(1m)2mx1=2m2mx1=22mm2mx1=1m12x2=1(1m)2mx2=11+m)2mx2=m2mx2=12

 

heureka Jul 13, 2015
 #5
avatar+14538 
+5

 x = 12      und      x = 1m12

 

Gruß radix !

 

 Jul 13, 2015
 #6
avatar+262 
0

Oh sorry i haven't seen his answer

 Jul 13, 2015
 #7
avatar+118696 
+5

Why didn't you see his answer?

The fact that you had an answer would have been clearly displayed in your watch list.  :/

 Jul 13, 2015
 #8
avatar+262 
0

yes but i think i confused it with other answer and reloaded the browser whithout noticed that its exactly the answer i need

 Jul 13, 2015
 #9
avatar+262 
+5

 i think its because my watchlist is a bit messy with all of my posts and i should remove some of them from my watchlist

 Jul 13, 2015
 #10
avatar+118696 
+5

Ok, it is not a problem sabi,  

I am sure you will get yourself better organised.   

 Jul 14, 2015

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